分块矩阵求逆法：副对角线形式（C010）

问题

已知，$\mathit{A}$ 和 $\mathit{B}$ 是元素 非 全 为 零 的方阵，$\mathit{O}$ 是元素 全 为 零 的方阵
则，根据可逆矩阵的性质，${\left(\begin{array}{ll}\mathit{O}& \mathit{A}\\ \mathit{B}& \mathit{O}\end{array}\right)}^{-1}$ $=$ $?$

选项

[A].   ${\left(\begin{array}{ll}\mathit{O}& \mathit{A}\\ \mathit{B}& \mathit{O}\end{array}\right)}^{-1}$ $=$ $\left(\begin{array}{cc}{\mathit{B}}^{-1}& \mathit{O}\\ \mathit{O}& {\mathit{A}}^{-1}\end{array}\right)$

[B].   ${\left(\begin{array}{ll}\mathit{O}& \mathit{A}\\ \mathit{B}& \mathit{O}\end{array}\right)}^{-1}$ $=$ $\left(\begin{array}{cc}\mathit{O}& {\mathit{A}}^{\mathbf{-}\mathbf{1}}\\ {\mathit{B}}^{\mathbf{-}\mathbf{1}}& \mathit{O}\end{array}\right)$

[C].   ${\left(\begin{array}{ll}\mathit{O}& \mathit{A}\\ \mathit{B}& \mathit{O}\end{array}\right)}^{-1}$ $=$ $\left(\begin{array}{cc}\mathit{O}& {\mathit{B}}^{\mathrm{\top }}\\ {\mathit{A}}^{\mathrm{\top }}& \mathit{O}\end{array}\right)$

[D].   ${\left(\begin{array}{ll}\mathit{O}& \mathit{A}\\ \mathit{B}& \mathit{O}\end{array}\right)}^{-1}$ $=$ $\left(\begin{array}{cc}\mathit{O}& {\mathit{B}}^{-1}\\ {\mathit{A}}^{-1}& \mathit{O}\end{array}\right)$

   答 案   

${\left(\begin{array}{ll}\mathit{O}& \mathit{A}\\ \mathit{B}& \mathit{O}\end{array}\right)}^{-1}$ $=$ $\left(\begin{array}{cc}\mathit{O}& {\mathit{B}}^{-1}\\ {\mathit{A}}^{-1}& \mathit{O}\end{array}\right)$